Positive decimal integers with the digit 1 occurring exactly twice
Find positive decimal integers n in which the digit 1 occurs exactly twice, where n < 1,000.
Positive decimal integers with the digit 1 occurring exactly twice is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
- Task
11l
L(n) 1..999
I String(n).count(‘1’) == 2
print(n, end' ‘ ’)- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
8080 Assembly
org 100h
loop: lda tho ; Are we there yet?
cpi '0'
rnz
call incr ; If not, increment
lxi b,0203h ; Need two ones, 3 steps
mvi a,'1'
lxi h,acc
ones: cmp m ; Count the ones
jnz $+4
dcr b
dcr c
inx h
jnz ones
xra a ; If two ones, print
ora b
cz prn
jmp loop
incr: lxi h,acc ; Increment accumulator
mvi a,'9'+1
incrl: inr m
cmp m
rnz
mvi m,'0'
inx h
jmp incrl
prn: lxi h,acc+4 ; Print accumulator w/o leading zero
mvi a,'0'
skip: dcx h
cmp m
jz skip
prl: push h
mvi c,2 ; CP/M print character
mov e,m
call 5
pop h
dcx h
xra a
cmp m
jnz prl
mvi c,9 ; CP/M print newline
lxi d,nl
jmp 5
acc: db '000' ; Accumulator (stored backwards)
tho: db '0' ; Thousands digit (stop if not 0 anymore)
nl: db 13,10,'$' ; Newline
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Action!
BYTE FUNC OnesCount(INT x)
BYTE c,d
c=0
WHILE x#0
DO
d=x MOD 10
IF d=1 THEN
c==+1
FI
x==/10
OD
RETURN (c)
PROC Main()
INT i
FOR i=0 TO 999
DO
IF OnesCount(i)=2 THEN
PrintI(i) Put(32)
FI
OD
RETURN- Output:
Screenshot from Atari 8-bit computer
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Ada
with Ada.Text_IO; use Ada.Text_IO;
procedure Positives_With_1_Twice is
function One_Twice (N : Positive) return Boolean is
NN : Natural := N;
One_Count : Natural := 0;
begin
while NN > 0 loop
if NN mod 10 = 1 then
One_Count := One_Count + 1;
end if;
NN := NN / 10;
end loop;
return One_Count = 2;
end One_Twice;
begin
for N in 1 .. 999 loop
if One_Twice (N) then
Put (N'Image); Put (" ");
end if;
end loop;
New_Line;
end Positives_With_1_Twice;
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
ALGOL 68
Generates the numbers. In order to print them in order, a table of double 1 numbers yes/no is generated.
BEGIN # find numbers where the digit 1 occurs twice, up to 999 #
[ 1 : 999 ]BOOL double 1; FOR i TO UPB double 1 DO double 1[ i ] := FALSE OD;
# generte the numbers #
FOR i FROM 0 TO 9 DO
IF i /= 1 THEN
double 1[ 110 + i ] := TRUE;
double 1[ 101 + ( i * 10 ) ] := TRUE;
double 1[ ( i * 100 ) + 11 ] := TRUE
FI
OD;
# print the numbers in order #
INT double 1 count := 0;
FOR i TO UPB double 1 DO
IF double 1[ i ] THEN
print( ( " ", whole( i, -3 ) ) );
IF ( double 1 count +:= 1 ) MOD 10 = 0 THEN print( ( newline ) ) FI
FI
OD
END- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
ALGOL W
Generates the numbers and sorts them into order using Sorting algorithms/Quicksort#ALGOL W.
begin % find numbers where the digit 1 occurs twice, up to 999 %
integer double1Count;
integer array double1 ( 1 :: 100 ); % assume there will be at most 100 numbers %
% Quicksorts in-place the array of integers v, from lb to ub - external %
procedure quicksort ( integer array v( * )
; integer value lb, ub
) ; algol "sortingAlgorithms_Quicksort" ;
% increments n by 1 and returns its new value %
integer procedure inc ( integer value result n ) ; begin n := n + 1; n end inc ;
% generate the numbers %
double1Count := 0;
for i := 0 until 9 do begin
if i not = 1 then begin
double1( inc( double1Count ) ) := 110 + i;
double1( inc( double1Count ) ) := 101 + ( i * 10 );
double1( inc( double1Count ) ) := ( i * 100 ) + 11
end if_i_ne_1
end for_i ;
% sort the numbers %
quickSort( double1, 1, double1Count );
% print the numbers %
for i := 1 until double1Count do writeon( i_w := 1, s_w := 1, double1( i ) );
write();
write( i_w := 1, s_w := 0, "Found ", double1Count, " numbers" )
end.- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 Found 27 numbers
APL
(⊢(/⍨)(2='1'+.=⍕)¨) ⍳1000
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Arturo
prints [11 101 110]
loop 2..9 'd -> prints ~"|d|11 1|d|1 11|d| "
- Output:
11 101 110 211 121 112 311 131 113 411 141 114 511 151 115 611 161 116 711 171 117 811 181 118 911 191 119
AWK
# syntax: GAWK -f NUMBERS_N_IN_WHICH_NUMBER_1_OCCUR_TWICE.AWK
BEGIN {
start = 1
stop = 999
for (i=start; i<=stop; i++) {
if (gsub(/1/,"&",i) == 2) {
printf("%4d%1s",i,++count%10?"":"\n")
}
}
printf("\nNumber 1 occurs twice %d-%d: %d\n",start,stop,count)
exit(0)
}
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 Number 1 occurs twice 1-999: 27
COBOL
IDENTIFICATION DIVISION.
PROGRAM-ID. TWO-ONES.
DATA DIVISION.
WORKING-STORAGE SECTION.
77 NUM PIC 9999.
77 FMT PIC ZZ9.
77 ONES PIC 9.
PROCEDURE DIVISION.
BEGIN.
PERFORM COUNT-ONES
VARYING NUM FROM 1 BY 1 UNTIL NUM IS EQUAL TO 1000.
STOP RUN.
COUNT-ONES.
MOVE ZERO TO ONES.
INSPECT NUM TALLYING ONES FOR ALL '1'.
IF ONES IS EQUAL TO 2,
MOVE NUM TO FMT,
DISPLAY FMT.
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Delphi
See Pascal
EasyLang
func dig1 n .
while n > 0
r += if n mod 10 = 1
n = n div 10
.
return r
.
for i to 999
if dig1 i = 2
write i & " "
.
.
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Euler
As with the Arturo, F# etc. samples, generates the sequence but doesn't sort it into order.
begin new dPos; new non1digits; label digitLoop;
non1digits <- ( 0, 2, 3, 4, 5, 6, 7, 8, 9 );
dPos <- 0;
digitLoop: if [ dPos <- dPos + 1 ] <= length non1digits then begin
new d;
d <- non1digits[ dPos ];
out [ d * 100 ] + 11; out [ d * 10 ] + 101; out 110 + d;
goto digitLoop
end else 0
end $
- Output:
NUMBER 11
NUMBER 101
NUMBER 110
NUMBER 211
NUMBER 121
NUMBER 112
NUMBER 311
NUMBER 131
NUMBER 113
NUMBER 411
NUMBER 141
NUMBER 114
NUMBER 511
NUMBER 151
NUMBER 115
NUMBER 611
NUMBER 161
NUMBER 116
NUMBER 711
NUMBER 171
NUMBER 117
NUMBER 811
NUMBER 181
NUMBER 118
NUMBER 911
NUMBER 191
NUMBER 119
F#
// 3 digit numbers with 2 ones. Nigel Galloway: July 6th., 2021
[0;2;3;4;5;6;7;8;9]|>List.collect(fun g->[[g;1;1];[1;g;1];[1;1;g]])|>List.iter(fun(n::g::l::_)->printf "%d " (n*100+g*10+l)); printfn ""
- Output:
11 101 110 211 121 112 311 131 113 411 141 114 511 151 115 611 161 116 711 171 117 811 181 118 911 191 119
Factor
USING: io math math.functions prettyprint sequences
sequences.extras ;
{ 0 2 3 4 5 6 7 8 9 }
[| n | { { n 1 1 } { 1 n 1 } { 1 1 n } } ] map-concat
[ <reversed> 0 [ 10^ * + ] reduce-index pprint bl ] each nl
- Output:
11 101 110 211 121 112 311 131 113 411 141 114 511 151 115 611 161 116 711 171 117 811 181 118 911 191 119
FreeBASIC
function numdig(byval n as integer, d as const integer) as uinteger
'counts the number of occurrences of digit d in the number n
dim as uinteger m = 0
while n
if n mod 10 = d then m+=1
n\=10
wend
return m
end function
for i as uinteger = 1 to 999
if numdig(i, 1) = 2 then print i;" ";
next i
print
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Free Pascal
See Pascal
Go
package main
import (
"fmt"
"rcu"
)
func main() {
fmt.Println("Decimal numbers under 1,000 whose digits include two 1's:")
var results []int
for i := 11; i <= 911; i++ {
digits := rcu.Digits(i, 10)
count := 0
for _, d := range digits {
if d == 1 {
count++
}
}
if count == 2 {
results = append(results, i)
}
}
for i, n := range results {
fmt.Printf("%5d", n)
if (i+1)%7 == 0 {
fmt.Println()
}
}
fmt.Println("\n\nFound", len(results), "such numbers.")
}
- Output:
Decimal numbers under 1,000 whose digits include two 1's: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 Found 27 such numbers.
J
Generating the numbers as the cartesian product of the eligible digits (and sorting the result, for "prettiness"):
~./:~10 #. >,{~.(2 1#1;i.10){~(! A.&i. ])3
11 101 110 111 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
jq
Works with gojq, the Go implementation of jq
This entry contains two solutions: the first is fast for this problem, but algorithmically inefficient; the second uses `countEquals` so candidates can be discarded quickly. The output is the same in both cases and so is shown only once.
Using `count`
def count(s): reduce s as $x (0; .+1);
range(1;1000)
| select( tostring | explode | 2 == count( select(.[] == 49))) # "1"Using `countEquals`
def countEquals($n; s):
label $out
| foreach (s, null) as $x (-1;
. + 1;
if $x == null then . == $n
elif . > $n then false, break $out
else empty
end);
range(1;1000)
| select( tostring
| countEquals(2; select(explode[] == 49))) # "1"- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Julia
totalddigisn(x, d = 1, n = 2; base=10) = count(j -> j == d, digits(x; base)) == n
println(filter(totalddigisn, 1:1000))
- Output:
[11, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 211, 311, 411, 511, 611, 711, 811, 911]
Mathematica / Wolfram Language
sol = Cases[Range[999], _?(Count[IntegerDigits@#, 1] == 2 &)];
Partition[sol, Max[FactorInteger[Length@sol][[All, 1]]]] // TableForm
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Ksh
#!/bin/ksh
# Positive decimal integers with the digit 1 occurring twice
# # Variables:
#
integer MAX=999
# # Functions:
#
######
# main #
######
for ((i=10; i<MAX; i++)); do
[[ ${i} == *{2}(1)* ]] && printf "%d " ${i}
done
echo
- Output:
11 110 111 112 113 114 115 116 117 118 119 211 311 411 511 611 711 811 911
MiniZinc
%Permutations with some identical elements. Nigel Galloway: July 6th., 2021
include "count.mzn";
array [1..3] of var 0..9: N; constraint count(N,1,2);
output [show((N[1]*100)+(N[2]*10)+N[3])]- Output:
110 ---------- 101 ---------- 11 ---------- 211 ---------- 311 ---------- 411 ---------- 511 ---------- 611 ---------- 711 ---------- 811 ---------- 911 ---------- 121 ---------- 131 ---------- 141 ---------- 151 ---------- 161 ---------- 171 ---------- 181 ---------- 191 ---------- 112 ---------- 113 ---------- 114 ---------- 115 ---------- 116 ---------- 117 ---------- 118 ---------- 119 ---------- ========== Finished in 206msec
Nim
import sugar, strutils
let result = collect(newSeq):
for n in 1..<1000:
if count($n, '1') == 2: n
echo "Found ", result.len, " numbers:"
echo result.join(" ")
- Output:
Found 27 numbers: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Pascal
program positiveDecimalIntegersWithTheDigit1occurringExactlyTwice(output);
var
n: integer;
begin
for n := 1 to 999 do
begin
if ord(n mod 10 = 1) + ord(n mod 100 div 10 = 1) + ord(n div 100 = 1) = 2 then
begin
writeLn(n)
end
end
end.
Perl
#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Numbers_n_in_which_number_1_occur_twice
use warnings;
my @twoones = grep tr/1// =~ 2, 1 .. 1000;
print "@twoones\n" =~ s/.{60}\K /\n/gr;
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Phix
function two_ones(string s) return length(find_all('1',s))=2 end function sequence res = filter(apply(tagset(999),sprint),two_ones) printf(1,"%d found:\n %s\n",{length(res),join_by(res,1,9," ","\n ")})
- Output:
27 found: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Pike
int main() {
int limit = 1000;
for(int i = 0; i < limit + 1; i++) {
if(String.count((string)i, "1") == 2) {
write((string)i + " ");
}
}
write("\n");
return 0;
}
- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
PL/M
... under CP/M (or an emulator)
100H: /* FIND NUMBERS BELOW 1000 WHERE 1 OCCURS TWICE IN THIER DECIMAL */
/* REPRESENTATION */
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* TASK */
DECLARE ( I, V, COUNT, ONE$COUNT ) ADDRESS;
COUNT = 0;
DO I = 10 TO 999;
V = I;
ONE$COUNT = 0;
DO WHILE V > 0;
IF V MOD 10 = 1 THEN DO;
ONE$COUNT = ONE$COUNT + 1;
END;
V = V / 10;
END;
IF ONE$COUNT = 2 THEN DO;
CALL PR$CHAR( ' ' );
IF I < 100 THEN CALL PR$CHAR( ' ' );
CALL PR$NUMBER( I );
IF ( COUNT := COUNT + 1 ) MOD 10 = 0 THEN CALL PR$NL;
END;
END;
EOF- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
PROMAL
;;; find numbers below 1000 where 1 occurs twice in their decimal representation
PROGRAM oneTwice
INCLUDE LIBRARY
WORD i
WORD v
WORD count
WORD oneCount
BEGIN
count = 0
FOR i = 10 TO 999
v = i
oneCount = 0
WHILE v > 0
IF v % 10 = 1
oneCount = oneCount + 1
v = v / 10
IF oneCount = 2
OUTPUT " #3I", i
count = count + 1
IF count % 10 = 0
OUTPUT "#C"
END- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
Python
#Aamrun, 5th October 2021
from itertools import permutations
for i in range(0,10):
if i!=1:
baseList = [1,1]
baseList.append(i)
[print(int(''.join(map(str,j)))) for j in sorted(set(permutations(baseList)))]
- Output:
11 101 110 112 121 211 113 131 311 114 141 411 115 151 511 116 161 611 117 171 711 118 181 811 119 191 911
Quackery
[ 0 swap
[ dup 0 != while
10 /mod 1 = if
[ dip 1+ ]
again ]
drop 2 = ] is two-1s ( n --> b )
[] 1000 times
[ i^ two-1s if
[ i^ join ] ]
echo- Output:
[ 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 ]
Raku
say display 10, '%3d', ^1000 .grep: { .comb.Bag{'1'} == 2 };
sub display {
cache $^c;
"{+$c} matching:\n" ~ $c.batch($^a)».fmt($^b).join: "\n"
}
- Yields:
27 matching: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911
REXX
version 1
Programming note: the three filters (marked below) could've been incorporated into one REXX statement if code golf is the desired goal.
/*REXX program finds positive decimal integers which contain exactly two ones (1s). */
parse arg hi cols . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi= 1000 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
w= 10 /*width of a number in any column. */
title= ' positive decimal integers which contain exactly two ones (1s) which are <' hi
say ' index │'center(title, 1 + cols*(w+1) )
say '───────┼'center("" , 1 + cols*(w+1), '─')
found= 0; idx= 1 /*initialize # integers and the index. */
$= /*a list of integers found (so far). */
do j=1 for hi-1 /*find positive integers within range. */
p= pos(1, j); if p==0 then iterate /*integer doesn't have a one (1)? Skip.*/ /* ◄■■■■■■■ a filter.*/
p= pos(1, j, p+1); if p==0 then iterate /* " " " a 2nd one? " */ /* ◄■■■■■■■ a filter.*/
p= pos(1, j, p+1); if p>0 then iterate /* " does " a 3rd one? " */ /* ◄■■■■■■■ a filter.*/
found= found + 1 /*bump the number of integers found. */
$= $ right(j, w) /*add an integer to the ──► $ list. */
if found//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
/*stick a fork in it, we're all done. */
if $\=='' then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/
say '───────┴'center("" , 1 + cols*(w+1), '─')
say
say 'Found ' found title
- output when using the default inputs:
index │ positive decimal integers which contain exactly two ones (1s) which are < 1000 ───────┼─────────────────────────────────────────────────────────────────────────────────────────────────────────────── 1 │ 11 101 110 112 113 114 115 116 117 118 11 │ 119 121 131 141 151 161 171 181 191 211 21 │ 311 411 511 611 711 811 911 ───────┴─────────────────────────────────────────────────────────────────────────────────────────────────────────────── Found 27 positive decimal integers which contain exactly two ones (1s) which are < 1000
version 2
Programming note: not all REXXes have the countstr BIF.
/*REXX program finds positive decimal integers which contain exactly two ones (1s). */
parse arg hi cols . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi= 1000 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
w= 10 /*width of a number in any column. */
title= ' positive decimal integers which contain exactly two ones (1s) which are <' hi
say ' index │'center(title, 1 + cols*(w+1) )
say '───────┼'center("" , 1 + cols*(w+1), '─')
found= 0; idx= 1 /*initialize # integers and the index. */
$= /*a list of integers found (so far). */
do j=1 for hi-1 /*find positive integers within range. */
if countstr(1, j)\==2 then iterate /*Doesn't have exactly 2 one's? Skip. */ /* ◄■■■■■■■ a filter.*/
found= found + 1 /*bump the number of integers found. */
$= $ right(j, w) /*add an integer to the ──► $ list. */
if found//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
/*stick a fork in it, we're all done. */
if $\=='' then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/
say '───────┴'center("" , 1 + cols*(w+1), '─')
say
say 'Found ' found title
- output is identical to the 1st REXX version.
Ring
load "stdlib.ring"
see "working..." + nl
see "Numbers n in which number 1 occur twice:" + nl
row = 0
sum = 0
limit = 1000
for n = 1 to limit
strn = string(n)
ind = count(strn,"1")
if ind = 2
see "" + n + " "
row++
if row%5 = 0
see nl
ok
ok
next
see nl + "Found " + row + " numbers" + nl
see "done..." + nl
func count(cstring,dstring)
sum = 0
while substr(cstring,dstring) > 0
sum++
cstring = substr(cstring,substr(cstring,dstring)+len(string(sum)))
end
return sum- Output:
working... Numbers n in which number 1 occur twice: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 Found 27 numbers done...
RPL
≪ →STR 0
1 3 PICK SIZE FOR j
OVER j DUP SUB "1" == + NEXT
SWAP DROP 2 ==
≫ 'ONLY2?' STO
≪ { }
1 1000 FOR j
IF j ONLY2? THEN j + END NEXT
≫ EVAL
- Output:
1: { 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 }
Ruby
p (1..1000).select{|n| n.digits.count(1) == 2}
- Output:
[11, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 211, 311, 411, 511, 611, 711, 811, 911]
SETL
program two_ones;
print([n : n in [1..999] | 2 = #[d : d in str n | val d = 1]]);
end program;- Output:
[11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911]
Sidef
say (1..1000 -> grep { .digits.count { _ == 1 } == 2 })
- Output:
[11, 101, 110, 112, 113, 114, 115, 116, 117, 118, 119, 121, 131, 141, 151, 161, 171, 181, 191, 211, 311, 411, 511, 611, 711, 811, 911]
Wren
import "./math" for Int
import "./fmt" for Fmt
System.print("Decimal numbers under 1,000 whose digits include two 1's:")
var results = (11..911).where { |i| Int.digits(i).count { |d| d == 1 } == 2 }.toList
Fmt.tprint("$5d", results, 7)
System.print("\nFound %(results.count) such numbers.")
- Output:
Decimal numbers under 1,000 whose digits include two 1's: 11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 Found 27 such numbers.
XPL0
func Ones(N); \Return count of 1's in N
int N, Count;
[Count:= 0;
repeat N:= N/10;
if rem(0) = 1 then Count:= Count+1;
until N = 0;
return Count;
];
int N, Count;
[for N:= 1 to 1000-1 do
if Ones(N) = 2 then
[IntOut(0, N);
Count:= Count+1;
if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
];
CrLf(0);
IntOut(0, Count);
Text(0, " such numbers found below 1000.
");
]- Output:
11 101 110 112 113 114 115 116 117 118 119 121 131 141 151 161 171 181 191 211 311 411 511 611 711 811 911 27 such numbers found below 1000.